There are 20 people around a circular table.We have to choose $3$ of them such that at least $2$ of them are sitting together.In how many ways can this be done?

Number of ways of choosing 3 people sitting adjacent to each other is 20.Number of ways of picking 2 people adjacent to each other is also 20.For each of those latter positions,there are 16 other people to choose from.

Therefore, $\text{Ans:} 20+20\cdot16=340$

In what other ways can this answer be derived?Can we use complementary counting somehow?Is there a better way to visualize what we are trying to compute?

2 Answers
2

We count the complement. This is less efficient than your way for $3$, but becomes useful for larger numbers.

We first count something else: The number of ways to choose a Committee Chair, and then choose $2$ people to join the Chair, with none of the three people sitting next to each other.

The Chair can be chosen in $20$ ways.

Then we cannot use the two people next to the Chair. That leaves $17$. Write down $15=17-2$ stars, like this:
$$\ast\quad \ast\quad\ast\quad\ast\quad\ast\quad\ast\quad \ast\quad\ast\quad\ast\quad\ast\quad\ast\quad \ast\quad\ast\quad\ast\quad\ast$$
This determines $16$ "gaps" (the $14$ ordinary gaps that one can see, plus the $2$ "endgaps"). We will choose $2$ of these gaps to put our chosen people into. This can be done in $\binom{16}{2}$ ways. That gives a total of $(20)\binom{16}{2}$ choices of Committee with Chair, that satisfy the separation condition.

Now let's count the chairless committees that satisfy the separation condition. Each chairless committee has been counted $3$ times, so the required number is
$$\frac{(20)\binom{16}{2}}{3}.$$

Remark: This idea works if we have $n$ people, and we want to choose $r$ of them so that no two are adjacent. When we remove the Chair and the two adjacents, we need to write down $(n-3)-(r-1)$ stars, which gives $n-r-1$ "gaps." We choose $r-1$ of them. The same reasoning leads to
$$\frac{(n)\binom{n-r-1}{r-1}}{r}.$$

We can generate these by choosing the people in position $1$ (pink), position $2$ (blue) and position $k \in \{3,4,\ldots,19\}$ (one of the orange circles), then rotating cyclically. Since there are no automorphisms under this group action, we find $20 \times 17=340$ choices.